Answer:
[tex]a_{avg}=1324.69*\frac{m}{s^{2} }[/tex]
Step-by-step explanation:
The average acceleration is given by:
[tex]a_{avg}=\frac{\Delta v}{\Delta t} =\frac{v_{f}-v_{0}}{\Delta t}[/tex]
where [tex]v_{0}[/tex] is the velocity of the ball an instant before hitting the ground and [tex]v_{f}[/tex] an instant after.
we can obtain [tex]v_{0}[/tex] by applying the following formula:
[tex]v_{0}^{2} =v'_{0}^{2} -2*g*\Delta y[/tex]
where [tex]v'_{0}=0[/tex] since the ball falls from rest, g is the acceleration of gravity and [tex]\Delta y=(0-4.23)*m=-4.23*m[/tex]
then:
[tex]v_{0}=\sqrt{v'_{0}^{2} -2*g*\Delta y}[/tex]
[tex]\pm v_{0}=\sqrt{0^{2} -2*9.8*(-4.23)}[/tex]
[tex]v_{0}=-9.1054*\frac{m}{s}[/tex] (note that here we select the negative result because of the downwards direction of the velocity)
Similarly, we calculate [tex]v_{f}[/tex] with the formula:
[tex]v_{f}^{2} =v'_{f}^{2} -2*g*\Delta y[/tex]
where [tex]v'_{f}=0[/tex] is the velocity at the maximum height, and [tex]\Delta y=(2.54-0)*m=2.54*m[/tex]
then:
[tex]v_{f}=\sqrt{2*9.8*2.54}[/tex]
[tex]v_{f}=7.0558*\frac{m}{s}[/tex]
Finally the average acceleration is:
[tex]a_{avg}=\frac{7.0558+9.1054}{12.2*10^{-3} }*\frac{m}{s^{2}}=1324.69* \frac{m}{s^{2}}[/tex]
